To prove continuity of $\varphi(p) = \int_X |f|^p\ d\mu$ on $E = \{p: \varphi(p) < \infty\}$ where $0 < p < \infty$ 
Suppose $f$ is a complex measurable function on $X$, $\mu$ is a positive measure on $X$, and
$$\varphi(p) ~=~ \int_X |f|^p \; d\mu \quad (0 < p < \infty)$$
Let $E :=\{ p : \varphi(p) < \infty\}$. Assume $\|f\|_\infty > 0$. Prove that $\log\varphi$ is convex in the interior of $E$ and that $\varphi$ is continuous on $E$.

I have proved that $\log\varphi$ is convex on $\operatorname{int}E$, using Holder's inequality. We know that log-convexity implies convexity, so $\varphi$ is also convex on $\operatorname{int}E$, and hence continuous on $\operatorname{int}E$. How do we establish the continuity of $\varphi$ on $E$, though? I am missing exactly the points in $E\setminus\operatorname{int}E$.
I know this question has been previously answered here, but I'm unable to understand the proof. Moreover, I am trying to take a different approach to prove continuity. Let $\{p_n\}_{n\in\mathbb N}$ be a sequence in $E$, such that $p_n\to p$ (may or may not be in $E$ - we don't know if $E$ is closed, right?). I wish to show $\varphi(p_n)\to \varphi(p)$, i.e. $$\lim_{n\to\infty} \int_X |f|^{p_n}\ d\mu = \int_X |f|^p \ d\mu$$
This looks like a possible application of Lebesgue's Dominated Convergence Theorem, but I could not find a dominating function yet.
I would appreciate any help, thanks a lot!
I have already seen this answer - it did not help much. The proof was too convoluted, and I needed more details. @OliverDiaz and @robjohn were able to provide convincing detailed arguments, hence this post.
 A: Since $E$ is a convex subset of $[0,\infty)$, $E$ is an interval (possibly open, half open, or closed). If $E$ has no interior then either $E$ is empty or $E$ is a singleton $\{p\}$ (for example, for $f(x)=\frac{1}{x(1+|\log x|)^2}\mathbb{1}_{(0,\infty)}(x)$, and $(X,\mu)=(\mathbb{R},\lambda)$, then $E=\{1\}$). In this case there is nothing else to prove.
If $E$ has interior, then $E$ could be of the form $(a, b)$, $(a,b]$, $[a,b)$, $[a,b]$, $(a,\infty)$, or $[a,\infty)$, where $0\leq a<b<\infty$. Hence, if $r$ is an endpoint of the interval $E$, then either $r$ is finite or  $r=\infty$.
Case $r$ is finite: consider the set $A=\{|f| > 1\}$. Then $|f|^p\xrightarrow{E\ni p\rightarrow r}|f|^r$ monotonically on $A$ and on $X\setminus A$. Applying monotone (or dominated convergence) gives
\begin{align}
\int_A|f|^p&\xrightarrow{E\ni p\rightarrow r}\int_A|f|^r\\
\int_{X\setminus A}|f|^p&\xrightarrow{E\ni p\rightarrow r}\int_{X\setminus A}|f|^r
\end{align}
From this, we obtain $\lim_{E\ni p\rightarrow r}\varphi(p)=\|f\|^r_r=\phi(r)$. Notice that if $r\notin E$, then $\varphi(r)=\infty$.
Case $r=\infty$: Although this  case is not necessary to study continuity of $\varphi$ on $E$, a few observations can be made.

*

*(1) $\lim_{p\rightarrow\infty}\|f\|_p=\lim_{p\rightarrow\infty}(\varphi(p))^{1/p} =\|f\|_\infty$;  a proof can be found here.

*(2) One can also see that $\lim_{p\rightarrow\infty}\varphi(p)$ exists in $[0,\infty]$. Consider $A$ as before and let  $C=\{|f|<1\}$, and $p_0\in E$. Then $|f|^p\mathbb{1}_C\leq |f|^{p_0}\mathbb{1}_C$ for all $p>p_0$. Another application of monotone convergence and dominated convergence gives
\begin{align}
\lim_{p\rightarrow\infty}\int_X |f|^p\,d\mu&=\lim_{p\rightarrow\infty}\Big(\int_C|f|^p\,d\mu + \mu(|f|=1) +\int_A|f|^p\,d\mu\Big)\\
&=\mu(|f|=1)+\infty\cdot\mu(\{|f|>1\})\end{align}
If $\mu(|f|>1)>0$, then $\|f\|_\infty>1$ and $\lim\limits_{E\ni p\rightarrow\infty}\varphi(p)=\infty$. If $\mu(|f|>1)=0$, then $\lim\limits_{E\ni p\rightarrow\infty}\varphi(p)=\mu(|f|=1)$.

Notice that unless $\|f\|_\infty<1$ or $\|f\|_\infty>1$, $\varphi(\infty)$ is not well defined.
A: Since $\varphi$ is log-convex, $E$ is an interval.

Least Upper Bound is Included
Suppose that $b$ is the least upper bound of $E$. If $b\not\in E$, there is nothing to prove, so suppose that $b\in E$. Let $b_n\in E$ and $b_n\to b$ monotonically (increasing).
Define
$$
Y=\{x\in X:|f(x)|\le1\}
$$
Since $b_n\le b$,
$$
\int_{X\setminus Y}|f(x)|^{b_n}\,\mathrm{d}\mu\le\int_{X\setminus Y}|f(x)|^b\,\mathrm{d}\mu\lt\infty
$$
and since $b_1\le b_n\le b$,
$$
\int_Y|f(x)|^b\,\mathrm{d}\mu\le\int_Y|f(x)|^{b_n}\,\mathrm{d}\mu\le\int_Y|f(x)|^{b_1}\,\mathrm{d}\mu\lt\infty
$$
Thus, $|f(x)|^b[x\in X\setminus Y]+|f(x)|^{b_1}[x\in Y]$ is a dominating function proving that
$$
\lim_{b_n\to b}\int_X|f(x)|^{b_n}\,\mathrm{d}\mu=\int_X|f(x)|^b\,\mathrm{d}\mu
$$

Greatest Lower Bound is Included
Suppose that $a$ is the greatest lower bound of $E$. If $a\not\in E$, there is nothing to prove, so suppose that $a\in E$. Let $a_n\in E$ and $a_n\to a$ monotonically (decreasing).
Define
$$
Y=\{x\in X:|f(x)|\le1\}
$$
Since $a_n\ge a$,
$$
\int_Y|f(x)|^{a_n}\,\mathrm{d}\mu\le\int_Y|f(x)|^a\,\mathrm{d}\mu\lt\infty
$$
and since $a_1\ge a_n\ge a$,
$$
\int_{X\setminus Y}|f(x)|^a\,\mathrm{d}\mu\le\int_{X\setminus Y}|f(x)|^{a_n}\,\mathrm{d}\mu\le\int_{X\setminus Y}|f(x)|^{a_1}\,\mathrm{d}\mu\lt\infty
$$
Thus, $|f(x)|^{a_1}[x\in X\setminus Y]+|f(x)|^a[x\in Y]$ is a dominating function proving that
$$
\lim_{a_n\to a}\int_X|f(x)|^{a_n}\,\mathrm{d}\mu=\int_X|f(x)|^a\,\mathrm{d}\mu
$$
